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Byju's Answer
Standard XII
Mathematics
Summation by Sigma Method
Given log a...
Question
Given
log
a
2
=
x
,
log
a
3
=
y
and
log
a
5
=
z
. Find the value in each of the following in terms of
x
,
y
and
z
.
(i)
log
a
15
(ii)
log
a
8
(iii)
log
a
30
(iv)
log
a
(
27
125
)
(v)
log
a
(
3
1
3
)
(vi)
log
a
1.5
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Solution
(i)
log
a
15
=
log
a
3
+
log
a
5
=
y
+
z
(ii)
log
a
8
=
log
a
2
3
=
3
log
a
2
=
3
x
(iii)
log
a
30
=
log
a
(
2
×
3
×
5
)
=
log
a
2
+
log
a
3
+
log
a
5
=
x
+
y
+
z
(iv)
log
a
(
27
125
)
=
log
a
(
3
5
)
3
=
3
log
a
3
−
3
log
a
5
=
3
(
y
−
z
)
(v)
log
a
2
+
log
a
5
−
log
a
3
=
x
+
z
−
y
(vi)
log
a
(
3
2
)
=
log
a
3
−
log
a
2
=
y
−
x
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Similar questions
Q.
State whether each of the following statements is true or false.
(i)
log
5
125
=
3
(ii)
log
1
2
8
=
3
(iii)
log
4
(
6
+
3
)
=
log
4
6
+
log
4
3
(iv)
log
2
(
25
3
)
=
log
2
25
log
2
3
(v)
log
1
3
3
=
−
1
(vi)
log
a
(
M
−
N
)
=
log
a
M
÷
log
a
N
Q.
Assertion :If
x
,
y
,
z
are positive and are connected by the relations:
log
3
x
+
log
9
y
+
log
9
z
=
2
log
4
x
+
log
2
y
+
log
4
z
=
2
log
25
x
+
log
25
y
+
log
5
z
=
2
,
then numerical value of
y
z
+
z
x
+
x
y
is
3731
900
Reason: If
x
,
y
are positive and
a
,
b
>
0
,
a
,
b
≠
1
, then
log
a
(
x
y
)
=
log
a
x
+
log
a
y
and
log
a
x
=
log
b
x
log
b
a
Q.
If
log
(
a
+
b
)
=
log
a
+
log
b
, find
a
in terms of
b
.
Q.
Based on this information answer the questions given
If
(
a
>
1
,
x
>
1
)
or
(
0
<
a
<
1
,
0
<
x
<
1
)
.
then
log
a
x
>
0
i.e.,
log
a
x
is positive.
If
(
0
<
a
<
1
,
x
>
1
)
or
(
a
>
1
,
0
<
x
<
1
)
, then
log
a
x
<
0
If
a
>
b
then
log
b
a
>
1
and
log
a
b
<
1.
Determine the sign of
log
a
(
3.162
)
log
a
(
2
/
3
)
.
Q.
If
c
(
a
−
b
)
=
a
(
b
−
c
)
then
log
(
a
+
c
)
−
log
(
a
−
2
b
+
c
)
log
(
a
−
c
)
equals
(Assume all terms are defined)
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